The library with $999$ books. 
In the town of Capibara there is a library with books in $999$ themes. Since Capibara is an international town they have books in various languages. We know that for every language we can find exactly three themes containing books in that language. We also know that for every pair of themes, there is exactly one language which contains books in both those themes.
Prove there is a set of $250$ languages such that every theme contains books in at most 1 of those $250$ languages.

Can this problem be transferred to a graph theory problem?
 A: If I'm understanding the scenario correctly, you could look at it as a problem related to a colouring of the edges of the complete graph $K_{999}$.
So let $K_{999}$ be the complete graph where the vertices $\{t_1,\ldots,t_{999}\}$ represent the themes. Let the "colours" be the languages $\{\ell_1,\ldots,\ell_m\}$ and colour edge $(t_i,t_j)$ with the unique language in which both themes $t_i$ and $t_j$ have books in that language. 
That every language's books fall into exactly one of three themes implies that for every $\ell_k$, the subgraph induced by the edges coloured $\ell_k$ is a triangle.
The problem is to show that one can find a set of $250$ monochromatic triangles, no two of which share a vertex.
Added: Here's a proof for your problem that I promised in the comment (adapted from the cited paper). I don't know if this is the "contest" solution.
So suppose we are in the situation above, i.e., we have a partition $\Omega$ of $E(K_{999})$ into triangles. Let $\Pi$ be a maximum-sized set of mutually vertex-disjoint triangles in $\Omega$. We want to show that $\alpha:=|\Pi|\geq 250$. 
Let $P$ be the vertices used by the triangles in $\Pi$ and Let $S$ be the subgraph induced by $V(K_{999})\setminus P$. We have $|P|=3\alpha$ and so $|V(S)|=999-3\alpha$.
Since $\Pi$ is of maximum size, $S$ contains no triangles from $\Omega$. Hence, for every edge in $S$, the third vertex of the triangle in $\Omega$ containing that edge is in $P$. Let $X\subseteq P$ be the set of all vertices in $P$ which form such a third vertex, and for $x\in X$, let $b(x)$ be the set of edges in $S$ for which $x$ is that third vertex. Then $\cup_{x\in X} b(x)$ is a partition of the edges in $S$ and so $${999-3\alpha\choose 2} = \sum_{x\in X} |b(x)|.$$
Next, note that the edges in any $b(x)$ are mutually vertex-disjoint and so $|b(x)|\leq \frac{999-3\alpha}{2}$.
On the other hand, suppose $\pi\in\Pi$ contains at least two vertices of $X$. In this case we claim that for $x\in \pi\cap $X, we have $|b(x)|\leq 2$. To see this, suppose $b(x)$ has at least three edges and let $y$ be another vertex in $\pi\cap X$. Let $e\in b(y)$. Then there must be an $e^\prime\in b(x)$ that is vertex-disjoint from $e$. Thus $y$ together with $e$ forms a triangle $\pi_0$ that is disjoint from the triangle $\pi_1$ formed by $x$ and $e^\prime$. But now $\left(\Pi\setminus\{\pi\}\right)\cup\{\pi_0,\pi_1\}$ is a set of mutually vertex-disjoint triangles with $\alpha+1$ triangles, contradicting the maximality of $\Pi$.
Hence,
\begin{align*} {999-3\alpha\choose 2} &=\sum_{x\in X} |b(x)|\\
&=\sum_{\pi\in \Pi}\sum_{x\in \pi\cap X} |b(x)|\\
&=\sum_{\substack{\pi\in\Pi:\\|\pi\cap X|\geq 2}}\sum_{x\in \pi\cap X} |b(x)| + \sum_{\substack{\pi\in\Pi:\\|\pi\cap X|\leq 1}}\sum_{x\in \pi\cap X} |b(x)|\\
&\leq 6A+\frac{999-3\alpha}{2}B,
\end{align*}
where $A$ is the number of $\pi\in \Pi$ with $|X\cap\pi|\geq 2$ and $B$ is the number of $\pi\in\Pi$ with $|X\cap\pi|\leq 1$. Of course, $A+B=\alpha$. 
If $\frac{999-3\alpha}{2}\leq 6$, then $329\leq\alpha$ so we're done. If $\frac{999-3\alpha}{2}>6$, then the above inequality extends to $${999-3\alpha\choose 2}< \left(\frac{999-3\alpha}{2}\right)\alpha,$$ which reduces to $249.5<\alpha$. Since $\alpha$ is an integer, we can conclude $250\leq \alpha$.
